We note first that division by zero is not allowed and that for this reason \(2x-6\) may not be equal to zero and that therefore \(x=3\) is not a solution.

We now distinguish two cases, namely \(2x-6>0\) and \(2x-6<0\).
In both cases we multiply the inequality on both sides by \(2x-6\) because we then get a linear inequality, for which we know there is a solution method.

Suppose \(2x-6>0\), i.e. \(x> 3\). Then we get \(3<-8(2x-6)\).
When we move everything with \(x\) to the left and all constant terms to the right, we get \(16x<45\).
Then, dvision by the coefficient of \(x\)gives \(x < {{45}\over{16}}\).
So we have the following system of inequalities: \(x> 3\,\wedge\; x < {{45}\over{16}}\)
and this simplifies to \(\text{an empty solution set}\).

Suppose \(2x-6<0\), i.e. \(x< 3\). Then we get \(3>-8(2x-6)\).
When we move everything with \(x\) to the left and all constant terms to the right, we get \(16x>45\).
Then, division by the coefficient of \(x\) gives \(x > {{45}\over{16}}\).
So we have the following system of inequalities: \(x< 3\,\wedge\; x > {{45}\over{16}}\)
and this simplifies to \({{45}\over{16}}\lt x\lt 3\).

The solution of the original inequality is \({{45}\over{16}}\lt x\lt 3\).